Question

1g of oleum sample is diluted with water. The solution required 54mL of 0.4N NaOH for complete neutralization. The $\%$ of free $S{O_3}$ in the sample is:
A.74
B.26
C.20
D.None of these

Answer 1

Oleum (Latin for "oil"), also known as fuming sulfuric acid, is a term that refers to sulfuric acid solutions containing varying compositions of sulphur trioxide, or more simply, disulfuric acid (also known as pyrosulfuric acid). The CAS number 8014-95-7 is assigned to oleum.

Complete answer: A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralisation. In a water reaction, neutralisation ensures that there are no excess hydrogen or hydroxide ions in the solution. The acid strength of the reactants determines the pH of the neutralised solution. The expression "neutralisation" is used in the sense of a chemical reaction between an acid and a base or alkali. This reaction was previously described as
Acid + base (alkali)$\to$ salt + water.
The number of ${H_2}S{O_4}$ equivalents is the same as the number of $S{O_3}$ equivalents. It's almost the same as the number of NaOH equivalents.
The mass of one counterpart, that is, the mass of a given substance that can blend with or displace a fixed quantity of another substance, is known as equivalent weight. The mass of an element that mixes with or displaces 1.008 grams of hydrogen, 8.0 grams of oxygen, or 35.5 grams of chlorine is its equivalent weight.
$\text{Equivalent mass} = \dfrac{{{\text{ Atomic mass }}}}{{{\text{ Replaceable }}{{\text{H}}^ + }{\text{/O}}{{\text{H}}^ - }{\text{atoms }}}}$
Now using the given formula
Equivalents of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$+ Equivalents of ${\text{S}}{{\text{O}}_3}$= Equivalents of ${\text{NaOH}}$
Molecular mass of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ = 98 g
Molecular mass of ${\text{S}}{{\text{O}}_3}$ = 80 g
Let the replaceable ${H^ + }/O{H^ - }$atoms be x
Upon substituting we get
$\dfrac{{\text{x}}}{{98}} \times 2 + \dfrac{{(1 - {\text{x}}) \times 2}}{{80}} = 54 \times 0.4 \times {10^{ - 3}}$
X = 0.74
The percentage of free ${\text{S}}{{\text{O}}_3} = \dfrac{{1 - 0.74}}{1} \times 100 = 26\%$

Note:
The formula of a substance can also be used to calculate its percent composition. The mass of each product in one mole of the compound is first calculated using the subscripts in the formula. This is compounded by 100 percent and divided by the compound's molar mass.